Answer : 2446 years.
Explanation :
Length of semi major axis is,
According to Kepler's third law, square of time period of an orbit is directly proportional to the cube of the semi major axis.
i.e
where G is gravitational constant
M is mass of sun,
So,
since,
So, orbital period is approximately 2446 years.
Answer:
B = 0.025T.
Explanation:
See attachment below please.
Answer:
Vf = 23 m/s
Explanation:
First we need to find the distance covered by the motorcycle 2 when it passes motorcycle 1. Using the uniform speed equation for motorcycle 1:
s₁ = v₁t₁
where,
s₁ = distance covered by motorcycle 1 = ?
v₁ = speed of motorcycle 1 = 6.5 m/s
t₁ = time = 10 s
Therefore,
s₁ = (6.5 m/s)(10 s)
s₁ = 65 m
Now, for the distance covered by motorcycle 2 at the meeting point. Since, the motorcycle started 50 m ahead of motorcycle 2. Therefore,
s₂ = s₁ + 50 m
s₂ = 65 m + 50 m
s₂ = 115 m
Now, using second equation of motion for motorcycle 2:
s₂ = Vi t + (1/2)at²
where,
Vi = initial velocity of motorcycle 2 = 0 m/s
Therefore,
115 m = (0 m/s)(10 s) + (1/2)(a)(10 s)²
a = 230 m/100 s²
a = 2.3 m/s²
Now, using 1st equation of motion:
Vf = Vi + at
Vf = 0 m/s + (2.3 m/s²)(10 s)
Vf = 23 m/s
<u>The electrical </u><u>power P</u><u> delivered to the </u><u>resistor </u><u>via the work done on the individual charges passing through it is</u><u> Power = VI</u>
What is power in physics and examples?
- Power is the rate at which work is done. The SI unit for power is the watt W, where 1 watt equals 1 joule/second (1W=1J/s).
- Because work is energy transfer, power is also the rate at which energy is expended.
- A 60-W light bulb, for example, expends 60 J of energy per second.
Let Length of wire L and register R .
Electric field of charge q
Force on charge q
F = E . q ⇒ V q /L
Work done in moving charge q distance L'
W = F L =
Power = dw/ dt
= [ dq/dt = I ]
the power,
Power = VI
Learn more about power
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Answer:
- 2.425 x 10^5 J
Explanation:
The gravitational potential energy between earth and the bock is given by
Where, G is the universal gravitational constant = 6.67 x 10^-11 Nm^2/kg^2
M is the mass of earth = 5.8 x 10^24 kg
m is the mass of block = 4 kg
r be the radius of earth = 6380 km = 6380 x 10^3 m
by substituting the values in the above expression, we get
U = - 2.425 x 10^5 J